Derive a general expression for the phasor response I_L and V_o

Thread Starter


Joined Oct 3, 2010
The circuit in the figure below is operating in the sinusoidal steady state with \(V_{s}(t)=V_{A} cos(\omega t)\). Derive a general expression for the phasor response \(I_{L}\) and the voltage \(V_{o}\).

Relevant equations:

\(I= \frac{V_{A}}{Z_{EQ}}\), where Z is the impedance and V is the volage.

My attempt:

I converted everything to the phasor ("frequency") domain. So the value of the inductor is now \(Z=j \omega L\). The resistors do not change.


\(Z_{EQ} = R + (j \omega L || R)\), where \(||\) represents "parallel." (i.e. \(j \omega L\) in parallel with \(R\).)

\(Z_{EQ} = R + \frac{j \omega LR}{j \omega L + R}\)

I stopped there because I realized I would be nowhere near close to the correct answer. Am I even on the right track with this logic first of all?

Actual answers:
\(I_{L}= \frac{V_{A}}{R+2j \omega L}\)
\(V_{L}=\frac{j \omega L V_{A}}{R+2j \omega L}\)


Joined Feb 16, 2010
You were headed in the right direction with that idea.

Converting to the frequency domain, you get:

Next you use a current divider between jwL and R:

Try the voltage yourself and don't give up prematurely because you think you might be wrong.